# 线性代数作业代写Linear Algebra代考

## 内积空间Inner product space代写

• 凸优化convex analysis
• 控制理论Control theory
• 数学方法Mathematical methods
• 优化理论 optimazation

## 线性代数Linear Algebra的历史

The procedure for solving simultaneous linear equations (using counting rods) is now known as Gaussian elimination and appears in Chapter 8 of the ancient Chinese mathematical texts. Chapter 8 of Nine Chapters on the Art of Mathematics: Rectangular Arrays. Its use is illustrated in eighteen problems, with two to five equations.

In Europe, systems of linear equations arose with the introduction of coordinates into geometry by René Descartes in 1637. Indeed, in this new geometry, now known as Cartesian geometry, lines and planes are represented by linear equations, and calculating their intersection is equivalent to solving a system of linear equations.

The first systematic approach to solving linear systems was the use of determinants, first considered by Leibniz in 1693, and in 1750 Gabriel Cramer used them to give explicit solutions to linear systems, now known as Cramer’s rules. Later, Gauss further described the elimination method, which was originally classified as an advance in geodesy.

## 线性代数Linear Algebra课后作业代写

If $X$ and $Y$ are vectors in $\mathbb{R}^{3}$, then
$$|X \cdot Y| \leq|X| \cdot|Y| .$$
Moreover if $X \neq 0$ and $Y \neq 0$, then
$$\begin{gathered} X \cdot Y=|X| \cdot|Y| \Leftrightarrow Y=t X, t>0, \ X \cdot Y=-|X| \cdot|Y| \Leftrightarrow Y=t X, t<0 . \end{gathered}$$ Proof. If $X=0$, then inequality $8.3$ is trivially true. So assume $X \neq 0$. Now if $t$ is any real number, by equation $8.2$, \begin{aligned} 0 \leq|t X-Y|^{2} &=|t X|^{2}-2(t X) \cdot Y+|Y|^{2} \\ &=t^{2}|X|^{2}-2(X \cdot Y) t+|Y|^{2} \\ &=a t^{2}-2 b t+c_{,} \end{aligned} where $a=|X|^{2}>0, b=X \cdot Y, c=|Y|^{2}$.
Hence
\begin{aligned} a\left(t^{2}-\frac{2 b}{a} t+\frac{c}{a}\right) & \geq 0 \ \left(t-\frac{b}{a}\right)^{2}+\frac{c a-b^{2}}{a^{2}} & \geq 0 . \end{aligned}
Substituting $t=b / a$ in the last inequality then gives
$$\frac{a c-b^{2}}{a^{2}} \geq 0$$
so
$$|b| \leq \sqrt{a c}=\sqrt{a} \sqrt{c}$$
and hence inequality $8.3$ follows.
To discuss equality in the Cauchy-Schwarz inequality, assume $X \neq 0$ and $Y \neq 0$.
Then if $X \cdot Y=|X| \cdot|Y|$, wẻ haveẻ for âll $t$
\begin{aligned} |t X-Y|^{2} &=t^{2}|X|^{2}-2 t X \cdot Y+|Y|^{2} \ &=t^{2}|X|^{2}-2 t|X| \cdot|Y|+|Y|^{2} \ &=|t X-Y|^{2} . \end{aligned}
Taking $t=|X| /|Y|$ then gives $|t X-Y|^{2}=0$ and hence $t X-Y=0$. Hence $Y=t X$, where $t>0$. The case $X \cdot Y=-|X| \cdot | Y$ is proved similarly.